Question

Yellow light of wavelength 6000Å produces fringes of width 0.8 mm in Young’s double slit experiment. If the source is replaced by another monochromatic source of wavelength 7500Å and the separation between the slits is doubled then the fringe width becomes :

• A. 0.1mm
• B. 0.8mm
• C. 4.3mm
• D. 1mm

Answer 1

Answer: (B)

0.8mm

Answer 2

: Fringe width in first case

$\beta_{1} = \frac{D\lambda_{1}}{d}$…….. (i)

Fringe width in second case

$\beta_{2} = \frac{D\lambda_{2}}{2d}.$ …… (ii)

Divided equation (ii) by (i)

$\therefore\frac{\beta_{2}}{\beta_{1}} = \frac{D\lambda_{2}/2d}{D\lambda_{1}/d} = \frac{1}{2}.\frac{\lambda_{2}}{\lambda_{1}}or\beta_{2} = \frac{1}{2}.\frac{\lambda_{2}}{\lambda_{1}}.\beta_{1}$

$= \frac{1}{2} \times \frac{7500Å}{6000Å} \times 0.8mm = 0.5mm$